Question 1

Find a general solution for $y\frac{{dy}}{{dx}} = x$

Accepted Answer

${y^2} - {x^2} = C$

Answer

${y^2} + {x^2} = C$

Answer

${x^2} - {y^2} = C$

Answer

${x^2} + {y^2} = C$

Question 2

The general solution for $x\frac{{dy}}{{dx}} = y$ is ______.

Accepted Answer

$y = Ax$

Answer

$y = {e^x} + C$

Answer

$y = A{e^x}$

Answer

$y = x + C$

Question 3

Find the particular solution for $\frac{{dy}}{{dx}} = \frac{x}{{y\left( {{x^2} + 1} \right)}}$ such that $y\left( 0 \right) =  - 2$

Accepted Answer

$y =  - \sqrt {\ln \left( {{x^2} + 1} \right) + 4} $

Answer

$y = \ln \left( {{x^2} + 1} \right) - 2$

Answer

$y = {e^{{x^2} + 1}} - {e^{x + 1}} - 2$

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